>Say the set S = {1,2,3,...,(n-2),(n-1)}.>Then if we squared each element x_i in S, we would get a new>set S' (with (n-1)/2 elements) whose elements would be>x_i)^2 mod n. Note we are assuming (n-1)/2 is odd, so>n = 3 mod 4. If we square each element in S'>((x_i)^2)^2 mod n we would end up with the same exact set S'>and so on.>>Do you agree with this statement?

Yes, assuming also that n is a prime. Actually, it still holdsif n is a product of an odd number of primes all congruent to3 (mod 4). By the way -- usually the letter p is used to emphasize that a variable is prime.

One of my main complaints with your "statements" is that you assume things not stated. Can't you make a complete statement? Why do I always have to do it for you?

Here's what you're claiming now:

Proposition:

If p is prime with p = 3 (mod 4), then the sets

{x^2 (mod p) | x = 1,2,3 ... (p-1)}

{x^4 (mod p) | x = 1,2,3 ... (p-1)}

are equal.

Remark:

The above claim is true and is easily proved, but that does _not_ yield a proof of your prior claim:

False Proposition:

If p is prime with p = 3 (mod 4), then for all integers x,ynot congruent to 0,1,-1 (mod p), the sets